K 1 and K 2 for dissociation of H 2 S are 4 - 10-3 and 1 - 10-5 - Calculate sulphide ion concentration in 0-1 MH 2 S solution-A- 10-4B- 10-8- 10-5D- 10-6

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Byju's Answer

Standard XII

Chemistry

Relative Lowering of Vapour Pressure

K 1 and K 2 f...

Question

K₁ and K₂ for dissociation of H₂S are 4 x 10^-3 and 1 x 10^-5. Calculate sulphide ion concentration in 0.1 MH₂S solution.

$10^{- 4}$

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$10^{- 8}$

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$10^{- 5}$

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$10^{- 6}$

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Solution

The correct option is C
$10^{- 5}$

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Similar questions

$K_{1}$ and $K_{2}$ for dissociation of $H_{2} S$ are $4 \times 10^{- 3}$ and $1 \times 10^{- 5}$ . Calculate sulphide ion concentration in $0.1$ $M H_{2} S$ solution.

Q. Find the concentration of

H^{+}

ions in an aqueous solution, which is saturated with

H_{2} S

(

0.1 M

) as well as

H_{2} C O_{3}

(

0.2 M

[Use data:

K_{1} = 10^{- 7}, K_{2} = 10^{- 14}

for

H_{2} S, K_{1} = 4 \times 10^{- 7}, K_{2} = 4 \times 10^{- 11}

for

H_{2} C O_{3}]

Q. A solution contains 0.1 M

H_{2} S

and 0.3 M HCl. Calculate the conc. of

S^{2 -}

and

H S^{-}

ions in solution, Given

K_{a_{1}}

and

K_{a_{2}}

for

H_{2} S

are

10^{- 7}

and

1.3 \times 10^{- 13}

respectively.

Q. The concentrations of

S^{2 -}

and

H S^{-}

ions in the solution consisting of

0.1 M H_{2} S

and

0.3 M H C l

, given that

K_{a_{1}} = 1.0 \times 10^{- 7}

and

K_{a_{2}} = 1.3 \times 10^{- 13}

for

H_{2} S

Q. An aqueous solution contains

0.10 M H_{2} S

and

0.20 M H C l

. If the equilibrium constants for the formation

H S^{-}

from

H_{2} S

1.0 \times 10^{- 7}

and that of

S^{2 -}

from ions

H S^{-}

1.2 \times 10^{- 13}

then the concentration of

S^{2 -}

ions in aqueous solution is:

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K1 and K2 for dissociation of H2S are 4 x 10-3 and 1 x 10-5. Calculate sulphide ion concentration in 0.1 MH2S solution.

The correct option is C 10−5

K₁ and K₂ for dissociation of H₂S are 4 x 10^-3 and 1 x 10^-5. Calculate sulphide ion concentration in 0.1 MH₂S solution.

The correct option is C
$10^{- 5}$